Prove/disprove if A(n) = Ω(f(n)) AND B(n) = Ω(g(n)) then A(n)*B(n) = Ω(f(n)*g(n))

Question

I tried like this if $ c_0 f(n)\le A(n)$ for all $n\ge n_0$ $ \land $ $c_1$*g(n) $\le$ B(n) for all $n\ge n_1$ then $c_0 f(n)B(n)≤A(n)B(n)$ for all $n≥max⁡(n_0,n_1)$, now since $c_1 g(n)≤B(n)$ for all $n≥n_1$ we can replace $B(n)$ for $c_1 g(n)$ so $c_0 c_1 f(n)g(n)≤A(n)B(n)$ for all $n≥max⁡(n_0,n_1)$ so $S(n)T(n)=Ω(f(n)g(n))$